\(\int \frac {1}{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \, dx\) [37]

Optimal result

Integrand size = 29, antiderivative size = 529 \[ \int \frac {1}{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \, dx=-\frac {d \text {arctanh}\left (\frac {\sqrt {2} c+\sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}+\sqrt {2} d x}{\sqrt [4]{c} \sqrt {c^{3/2}-\sqrt {c^3+4 a d^2}}}\right )}{2 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}-\sqrt {c^3+4 a d^2}}}+\frac {d \text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}-\sqrt {2} (c+d x)}{\sqrt [4]{c} \sqrt {c^{3/2}-\sqrt {c^3+4 a d^2}}}\right )}{2 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}-\sqrt {c^3+4 a d^2}}}-\frac {d \log \left (\sqrt {c} \sqrt {c^3+4 a d^2}-\sqrt {2} \sqrt [4]{c} d \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}} \left (\frac {c}{d}+x\right )+d^2 \left (\frac {c}{d}+x\right )^2\right )}{4 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}}+\frac {d \log \left (\sqrt {c} \sqrt {c^3+4 a d^2}+\sqrt {2} \sqrt [4]{c} d \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}} \left (\frac {c}{d}+x\right )+d^2 \left (\frac {c}{d}+x\right )^2\right )}{4 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}} \]

[Out]

-1/4*d*arctanh((c*2^(1/2)+d*x*2^(1/2)+c^(1/4)*(c^(3/2)+(4*a*d^2+c^3)^(1/2))^(1/2))/c^(1/4)/(c^(3/2)-(4*a*d^2+c
^3)^(1/2))^(1/2))/c^(3/4)*2^(1/2)/(4*a*d^2+c^3)^(1/2)/(c^(3/2)-(4*a*d^2+c^3)^(1/2))^(1/2)+1/4*d*arctanh((-(d*x
+c)*2^(1/2)+c^(1/4)*(c^(3/2)+(4*a*d^2+c^3)^(1/2))^(1/2))/c^(1/4)/(c^(3/2)-(4*a*d^2+c^3)^(1/2))^(1/2))/c^(3/4)*
2^(1/2)/(4*a*d^2+c^3)^(1/2)/(c^(3/2)-(4*a*d^2+c^3)^(1/2))^(1/2)-1/8*d*ln(d^2*(c/d+x)^2+c^(1/2)*(4*a*d^2+c^3)^(
1/2)-c^(1/4)*d*(c/d+x)*2^(1/2)*(c^(3/2)+(4*a*d^2+c^3)^(1/2))^(1/2))/c^(3/4)*2^(1/2)/(4*a*d^2+c^3)^(1/2)/(c^(3/
2)+(4*a*d^2+c^3)^(1/2))^(1/2)+1/8*d*ln(d^2*(c/d+x)^2+c^(1/2)*(4*a*d^2+c^3)^(1/2)+c^(1/4)*d*(c/d+x)*2^(1/2)*(c^
(3/2)+(4*a*d^2+c^3)^(1/2))^(1/2))/c^(3/4)*2^(1/2)/(4*a*d^2+c^3)^(1/2)/(c^(3/2)+(4*a*d^2+c^3)^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 529, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1120, 1108, 648, 632, 212, 642} \[ \int \frac {1}{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \, dx=-\frac {d \text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt {\sqrt {4 a d^2+c^3}+c^{3/2}}+\sqrt {2} c+\sqrt {2} d x}{\sqrt [4]{c} \sqrt {c^{3/2}-\sqrt {4 a d^2+c^3}}}\right )}{2 \sqrt {2} c^{3/4} \sqrt {4 a d^2+c^3} \sqrt {c^{3/2}-\sqrt {4 a d^2+c^3}}}+\frac {d \text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt {\sqrt {4 a d^2+c^3}+c^{3/2}}-\sqrt {2} (c+d x)}{\sqrt [4]{c} \sqrt {c^{3/2}-\sqrt {4 a d^2+c^3}}}\right )}{2 \sqrt {2} c^{3/4} \sqrt {4 a d^2+c^3} \sqrt {c^{3/2}-\sqrt {4 a d^2+c^3}}}-\frac {d \log \left (\sqrt {c} \sqrt {4 a d^2+c^3}-\sqrt {2} \sqrt [4]{c} d \sqrt {\sqrt {4 a d^2+c^3}+c^{3/2}} \left (\frac {c}{d}+x\right )+d^2 \left (\frac {c}{d}+x\right )^2\right )}{4 \sqrt {2} c^{3/4} \sqrt {4 a d^2+c^3} \sqrt {\sqrt {4 a d^2+c^3}+c^{3/2}}}+\frac {d \log \left (\sqrt {c} \sqrt {4 a d^2+c^3}+\sqrt {2} \sqrt [4]{c} d \sqrt {\sqrt {4 a d^2+c^3}+c^{3/2}} \left (\frac {c}{d}+x\right )+d^2 \left (\frac {c}{d}+x\right )^2\right )}{4 \sqrt {2} c^{3/4} \sqrt {4 a d^2+c^3} \sqrt {\sqrt {4 a d^2+c^3}+c^{3/2}}} \]

[In]

Int[(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^(-1),x]

[Out]

-1/2*(d*ArcTanh[(Sqrt[2]*c + c^(1/4)*Sqrt[c^(3/2) + Sqrt[c^3 + 4*a*d^2]] + Sqrt[2]*d*x)/(c^(1/4)*Sqrt[c^(3/2)
- Sqrt[c^3 + 4*a*d^2]])])/(Sqrt[2]*c^(3/4)*Sqrt[c^3 + 4*a*d^2]*Sqrt[c^(3/2) - Sqrt[c^3 + 4*a*d^2]]) + (d*ArcTa
nh[(c^(1/4)*Sqrt[c^(3/2) + Sqrt[c^3 + 4*a*d^2]] - Sqrt[2]*(c + d*x))/(c^(1/4)*Sqrt[c^(3/2) - Sqrt[c^3 + 4*a*d^
2]])])/(2*Sqrt[2]*c^(3/4)*Sqrt[c^3 + 4*a*d^2]*Sqrt[c^(3/2) - Sqrt[c^3 + 4*a*d^2]]) - (d*Log[Sqrt[c]*Sqrt[c^3 +
4*a*d^2] - Sqrt[2]*c^(1/4)*d*Sqrt[c^(3/2) + Sqrt[c^3 + 4*a*d^2]]*(c/d + x) + d^2*(c/d + x)^2])/(4*Sqrt[2]*c^(
3/4)*Sqrt[c^3 + 4*a*d^2]*Sqrt[c^(3/2) + Sqrt[c^3 + 4*a*d^2]]) + (d*Log[Sqrt[c]*Sqrt[c^3 + 4*a*d^2] + Sqrt[2]*c
^(1/4)*d*Sqrt[c^(3/2) + Sqrt[c^3 + 4*a*d^2]]*(c/d + x) + d^2*(c/d + x)^2])/(4*Sqrt[2]*c^(3/4)*Sqrt[c^3 + 4*a*d
^2]*Sqrt[c^(3/2) + Sqrt[c^3 + 4*a*d^2]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1108

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}
, Dist[1/(2*c*q*r), Int[(r - x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(r + x)/(q + r*x + x^2), x], x
]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]

Rule 1120

Int[(P4_)^(p_), x_Symbol] :> With[{a = Coeff[P4, x, 0], b = Coeff[P4, x, 1], c = Coeff[P4, x, 2], d = Coeff[P4
, x, 3], e = Coeff[P4, x, 4]}, Subst[Int[SimplifyIntegrand[(a + d^4/(256*e^3) - b*(d/(8*e)) + (c - 3*(d^2/(8*e
)))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2, 0] && NeQ[d, 0]] /; FreeQ[p, x] &&
PolyQ[P4, x, 4] && NeQ[p, 2] && NeQ[p, 3]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{c \left (4 a+\frac {c^3}{d^2}\right )-2 c^2 x^2+d^2 x^4} \, dx,x,\frac {c}{d}+x\right ) \\ & = \frac {d \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}}{d}-x}{\frac {\sqrt {c} \sqrt {c^3+4 a d^2}}{d^2}-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}} x}{d}+x^2} \, dx,x,\frac {c}{d}+x\right )}{2 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}}+\frac {d \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}}{d}+x}{\frac {\sqrt {c} \sqrt {c^3+4 a d^2}}{d^2}+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}} x}{d}+x^2} \, dx,x,\frac {c}{d}+x\right )}{2 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}} \\ & = \frac {\text {Subst}\left (\int \frac {1}{\frac {\sqrt {c} \sqrt {c^3+4 a d^2}}{d^2}-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}} x}{d}+x^2} \, dx,x,\frac {c}{d}+x\right )}{4 \sqrt {c} \sqrt {c^3+4 a d^2}}+\frac {\text {Subst}\left (\int \frac {1}{\frac {\sqrt {c} \sqrt {c^3+4 a d^2}}{d^2}+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}} x}{d}+x^2} \, dx,x,\frac {c}{d}+x\right )}{4 \sqrt {c} \sqrt {c^3+4 a d^2}}-\frac {d \text {Subst}\left (\int \frac {-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}}{d}+2 x}{\frac {\sqrt {c} \sqrt {c^3+4 a d^2}}{d^2}-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}} x}{d}+x^2} \, dx,x,\frac {c}{d}+x\right )}{4 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}}+\frac {d \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}}{d}+2 x}{\frac {\sqrt {c} \sqrt {c^3+4 a d^2}}{d^2}+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}} x}{d}+x^2} \, dx,x,\frac {c}{d}+x\right )}{4 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}} \\ & = -\frac {d \log \left (\sqrt {c} \sqrt {c^3+4 a d^2}-\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}} (c+d x)+(c+d x)^2\right )}{4 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}}+\frac {d \log \left (\sqrt {c} \sqrt {c^3+4 a d^2}+\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}} (c+d x)+(c+d x)^2\right )}{4 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}}-\frac {\text {Subst}\left (\int \frac {1}{\frac {2 \sqrt {c} \left (c^{3/2}-\sqrt {c^3+4 a d^2}\right )}{d^2}-x^2} \, dx,x,-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}}{d}+2 \left (\frac {c}{d}+x\right )\right )}{2 \sqrt {c} \sqrt {c^3+4 a d^2}}-\frac {\text {Subst}\left (\int \frac {1}{\frac {2 \sqrt {c} \left (c^{3/2}-\sqrt {c^3+4 a d^2}\right )}{d^2}-x^2} \, dx,x,\frac {\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}}{d}+2 \left (\frac {c}{d}+x\right )\right )}{2 \sqrt {c} \sqrt {c^3+4 a d^2}} \\ & = -\frac {d \tanh ^{-1}\left (\frac {\sqrt [4]{c} \left (\sqrt {2} c^{3/4}-\sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}\right )+\sqrt {2} d x}{\sqrt [4]{c} \sqrt {c^{3/2}-\sqrt {c^3+4 a d^2}}}\right )}{2 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}-\sqrt {c^3+4 a d^2}}}-\frac {d \tanh ^{-1}\left (\frac {\sqrt [4]{c} \left (\sqrt {2} c^{3/4}+\sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}\right )+\sqrt {2} d x}{\sqrt [4]{c} \sqrt {c^{3/2}-\sqrt {c^3+4 a d^2}}}\right )}{2 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}-\sqrt {c^3+4 a d^2}}}-\frac {d \log \left (\sqrt {c} \sqrt {c^3+4 a d^2}-\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}} (c+d x)+(c+d x)^2\right )}{4 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}}+\frac {d \log \left (\sqrt {c} \sqrt {c^3+4 a d^2}+\sqrt {2} \sqrt [4]{c} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}} (c+d x)+(c+d x)^2\right )}{4 \sqrt {2} c^{3/4} \sqrt {c^3+4 a d^2} \sqrt {c^{3/2}+\sqrt {c^3+4 a d^2}}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.13 \[ \int \frac {1}{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \, dx=\frac {1}{4} \text {RootSum}\left [4 a c+4 c^2 \text {$\#$1}^2+4 c d \text {$\#$1}^3+d^2 \text {$\#$1}^4\&,\frac {\log (x-\text {$\#$1})}{2 c^2 \text {$\#$1}+3 c d \text {$\#$1}^2+d^2 \text {$\#$1}^3}\&\right ] \]

[In]

Integrate[(4*a*c + 4*c^2*x^2 + 4*c*d*x^3 + d^2*x^4)^(-1),x]

[Out]

RootSum[4*a*c + 4*c^2*#1^2 + 4*c*d*#1^3 + d^2*#1^4 & , Log[x - #1]/(2*c^2*#1 + 3*c*d*#1^2 + d^2*#1^3) & ]/4

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.12

method	result	size
default	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (d^{2} \textit {\_Z}^{4}+4 c d \,\textit {\_Z}^{3}+4 c^{2} \textit {\_Z}^{2}+4 a c \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3} d^{2}+3 \textit {\_R}^{2} c d +2 \textit {\_R} \,c^{2}}\right )}{4}\)	\(64\)
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (d^{2} \textit {\_Z}^{4}+4 c d \,\textit {\_Z}^{3}+4 c^{2} \textit {\_Z}^{2}+4 a c \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3} d^{2}+3 \textit {\_R}^{2} c d +2 \textit {\_R} \,c^{2}}\right )}{4}\)	\(64\)

[In]

int(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c),x,method=_RETURNVERBOSE)

[Out]

1/4*sum(1/(_R^3*d^2+3*_R^2*c*d+2*_R*c^2)*ln(x-_R),_R=RootOf(_Z^4*d^2+4*_Z^3*c*d+4*_Z^2*c^2+4*a*c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 905 vs. \(2 (408) = 816\).

Time = 0.29 (sec) , antiderivative size = 905, normalized size of antiderivative = 1.71 \[ \int \frac {1}{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \, dx=\frac {1}{8} \, \sqrt {-\frac {2 \, {\left (a c^{3} + 4 \, a^{2} d^{2}\right )} \sqrt {-\frac {d^{2}}{a c^{9} + 8 \, a^{2} c^{6} d^{2} + 16 \, a^{3} c^{3} d^{4}}} + 1}{a c^{3} + 4 \, a^{2} d^{2}}} \log \left (d^{2} x + c d + {\left (2 \, a c d^{2} + {\left (a c^{7} + 4 \, a^{2} c^{4} d^{2}\right )} \sqrt {-\frac {d^{2}}{a c^{9} + 8 \, a^{2} c^{6} d^{2} + 16 \, a^{3} c^{3} d^{4}}}\right )} \sqrt {-\frac {2 \, {\left (a c^{3} + 4 \, a^{2} d^{2}\right )} \sqrt {-\frac {d^{2}}{a c^{9} + 8 \, a^{2} c^{6} d^{2} + 16 \, a^{3} c^{3} d^{4}}} + 1}{a c^{3} + 4 \, a^{2} d^{2}}}\right ) - \frac {1}{8} \, \sqrt {-\frac {2 \, {\left (a c^{3} + 4 \, a^{2} d^{2}\right )} \sqrt {-\frac {d^{2}}{a c^{9} + 8 \, a^{2} c^{6} d^{2} + 16 \, a^{3} c^{3} d^{4}}} + 1}{a c^{3} + 4 \, a^{2} d^{2}}} \log \left (d^{2} x + c d - {\left (2 \, a c d^{2} + {\left (a c^{7} + 4 \, a^{2} c^{4} d^{2}\right )} \sqrt {-\frac {d^{2}}{a c^{9} + 8 \, a^{2} c^{6} d^{2} + 16 \, a^{3} c^{3} d^{4}}}\right )} \sqrt {-\frac {2 \, {\left (a c^{3} + 4 \, a^{2} d^{2}\right )} \sqrt {-\frac {d^{2}}{a c^{9} + 8 \, a^{2} c^{6} d^{2} + 16 \, a^{3} c^{3} d^{4}}} + 1}{a c^{3} + 4 \, a^{2} d^{2}}}\right ) + \frac {1}{8} \, \sqrt {\frac {2 \, {\left (a c^{3} + 4 \, a^{2} d^{2}\right )} \sqrt {-\frac {d^{2}}{a c^{9} + 8 \, a^{2} c^{6} d^{2} + 16 \, a^{3} c^{3} d^{4}}} - 1}{a c^{3} + 4 \, a^{2} d^{2}}} \log \left (d^{2} x + c d + {\left (2 \, a c d^{2} - {\left (a c^{7} + 4 \, a^{2} c^{4} d^{2}\right )} \sqrt {-\frac {d^{2}}{a c^{9} + 8 \, a^{2} c^{6} d^{2} + 16 \, a^{3} c^{3} d^{4}}}\right )} \sqrt {\frac {2 \, {\left (a c^{3} + 4 \, a^{2} d^{2}\right )} \sqrt {-\frac {d^{2}}{a c^{9} + 8 \, a^{2} c^{6} d^{2} + 16 \, a^{3} c^{3} d^{4}}} - 1}{a c^{3} + 4 \, a^{2} d^{2}}}\right ) - \frac {1}{8} \, \sqrt {\frac {2 \, {\left (a c^{3} + 4 \, a^{2} d^{2}\right )} \sqrt {-\frac {d^{2}}{a c^{9} + 8 \, a^{2} c^{6} d^{2} + 16 \, a^{3} c^{3} d^{4}}} - 1}{a c^{3} + 4 \, a^{2} d^{2}}} \log \left (d^{2} x + c d - {\left (2 \, a c d^{2} - {\left (a c^{7} + 4 \, a^{2} c^{4} d^{2}\right )} \sqrt {-\frac {d^{2}}{a c^{9} + 8 \, a^{2} c^{6} d^{2} + 16 \, a^{3} c^{3} d^{4}}}\right )} \sqrt {\frac {2 \, {\left (a c^{3} + 4 \, a^{2} d^{2}\right )} \sqrt {-\frac {d^{2}}{a c^{9} + 8 \, a^{2} c^{6} d^{2} + 16 \, a^{3} c^{3} d^{4}}} - 1}{a c^{3} + 4 \, a^{2} d^{2}}}\right ) \]

[In]

integrate(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c),x, algorithm="fricas")

[Out]

1/8*sqrt(-(2*(a*c^3 + 4*a^2*d^2)*sqrt(-d^2/(a*c^9 + 8*a^2*c^6*d^2 + 16*a^3*c^3*d^4)) + 1)/(a*c^3 + 4*a^2*d^2))
*log(d^2*x + c*d + (2*a*c*d^2 + (a*c^7 + 4*a^2*c^4*d^2)*sqrt(-d^2/(a*c^9 + 8*a^2*c^6*d^2 + 16*a^3*c^3*d^4)))*s
qrt(-(2*(a*c^3 + 4*a^2*d^2)*sqrt(-d^2/(a*c^9 + 8*a^2*c^6*d^2 + 16*a^3*c^3*d^4)) + 1)/(a*c^3 + 4*a^2*d^2))) - 1
/8*sqrt(-(2*(a*c^3 + 4*a^2*d^2)*sqrt(-d^2/(a*c^9 + 8*a^2*c^6*d^2 + 16*a^3*c^3*d^4)) + 1)/(a*c^3 + 4*a^2*d^2))*
log(d^2*x + c*d - (2*a*c*d^2 + (a*c^7 + 4*a^2*c^4*d^2)*sqrt(-d^2/(a*c^9 + 8*a^2*c^6*d^2 + 16*a^3*c^3*d^4)))*sq
rt(-(2*(a*c^3 + 4*a^2*d^2)*sqrt(-d^2/(a*c^9 + 8*a^2*c^6*d^2 + 16*a^3*c^3*d^4)) + 1)/(a*c^3 + 4*a^2*d^2))) + 1/
8*sqrt((2*(a*c^3 + 4*a^2*d^2)*sqrt(-d^2/(a*c^9 + 8*a^2*c^6*d^2 + 16*a^3*c^3*d^4)) - 1)/(a*c^3 + 4*a^2*d^2))*lo
g(d^2*x + c*d + (2*a*c*d^2 - (a*c^7 + 4*a^2*c^4*d^2)*sqrt(-d^2/(a*c^9 + 8*a^2*c^6*d^2 + 16*a^3*c^3*d^4)))*sqrt
((2*(a*c^3 + 4*a^2*d^2)*sqrt(-d^2/(a*c^9 + 8*a^2*c^6*d^2 + 16*a^3*c^3*d^4)) - 1)/(a*c^3 + 4*a^2*d^2))) - 1/8*s
qrt((2*(a*c^3 + 4*a^2*d^2)*sqrt(-d^2/(a*c^9 + 8*a^2*c^6*d^2 + 16*a^3*c^3*d^4)) - 1)/(a*c^3 + 4*a^2*d^2))*log(d
^2*x + c*d - (2*a*c*d^2 - (a*c^7 + 4*a^2*c^4*d^2)*sqrt(-d^2/(a*c^9 + 8*a^2*c^6*d^2 + 16*a^3*c^3*d^4)))*sqrt((2
*(a*c^3 + 4*a^2*d^2)*sqrt(-d^2/(a*c^9 + 8*a^2*c^6*d^2 + 16*a^3*c^3*d^4)) - 1)/(a*c^3 + 4*a^2*d^2)))

Sympy [A] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.17 \[ \int \frac {1}{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \, dx=\operatorname {RootSum} {\left (t^{4} \cdot \left (16384 a^{3} c^{3} d^{2} + 4096 a^{2} c^{6}\right ) + 128 t^{2} a c^{3} + 1, \left ( t \mapsto t \log {\left (x + \frac {- 1024 t^{3} a^{2} c^{4} d^{2} - 256 t^{3} a c^{7} + 16 t a c d^{2} - 4 t c^{4} + c d}{d^{2}} \right )} \right )\right )} \]

[In]

integrate(1/(d**2*x**4+4*c*d*x**3+4*c**2*x**2+4*a*c),x)

[Out]

RootSum(_t**4*(16384*a**3*c**3*d**2 + 4096*a**2*c**6) + 128*_t**2*a*c**3 + 1, Lambda(_t, _t*log(x + (-1024*_t*
*3*a**2*c**4*d**2 - 256*_t**3*a*c**7 + 16*_t*a*c*d**2 - 4*_t*c**4 + c*d)/d**2)))

Maxima [F]

\[ \int \frac {1}{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \, dx=\int { \frac {1}{d^{2} x^{4} + 4 \, c d x^{3} + 4 \, c^{2} x^{2} + 4 \, a c} \,d x } \]

[In]

integrate(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c),x, algorithm="maxima")

[Out]

integrate(1/(d^2*x^4 + 4*c*d*x^3 + 4*c^2*x^2 + 4*a*c), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 603, normalized size of antiderivative = 1.14 \[ \int \frac {1}{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \, dx=-\frac {\log \left (x + \sqrt {\frac {c^{2} d^{2} + 2 \, \sqrt {-a c} d^{3}}{d^{4}}} + \frac {c}{d}\right )}{4 \, {\left (d^{2} {\left (\sqrt {\frac {c^{2} d^{2} + 2 \, \sqrt {-a c} d^{3}}{d^{4}}} + \frac {c}{d}\right )}^{3} - 3 \, c d {\left (\sqrt {\frac {c^{2} d^{2} + 2 \, \sqrt {-a c} d^{3}}{d^{4}}} + \frac {c}{d}\right )}^{2} + 2 \, c^{2} {\left (\sqrt {\frac {c^{2} d^{2} + 2 \, \sqrt {-a c} d^{3}}{d^{4}}} + \frac {c}{d}\right )}\right )}} + \frac {\log \left (x - \sqrt {\frac {c^{2} d^{2} + 2 \, \sqrt {-a c} d^{3}}{d^{4}}} + \frac {c}{d}\right )}{4 \, {\left (d^{2} {\left (\sqrt {\frac {c^{2} d^{2} + 2 \, \sqrt {-a c} d^{3}}{d^{4}}} - \frac {c}{d}\right )}^{3} + 3 \, c d {\left (\sqrt {\frac {c^{2} d^{2} + 2 \, \sqrt {-a c} d^{3}}{d^{4}}} - \frac {c}{d}\right )}^{2} + 2 \, c^{2} {\left (\sqrt {\frac {c^{2} d^{2} + 2 \, \sqrt {-a c} d^{3}}{d^{4}}} - \frac {c}{d}\right )}\right )}} - \frac {\log \left (x + \sqrt {\frac {c^{2} d^{2} - 2 \, \sqrt {-a c} d^{3}}{d^{4}}} + \frac {c}{d}\right )}{4 \, {\left (d^{2} {\left (\sqrt {\frac {c^{2} d^{2} - 2 \, \sqrt {-a c} d^{3}}{d^{4}}} + \frac {c}{d}\right )}^{3} - 3 \, c d {\left (\sqrt {\frac {c^{2} d^{2} - 2 \, \sqrt {-a c} d^{3}}{d^{4}}} + \frac {c}{d}\right )}^{2} + 2 \, c^{2} {\left (\sqrt {\frac {c^{2} d^{2} - 2 \, \sqrt {-a c} d^{3}}{d^{4}}} + \frac {c}{d}\right )}\right )}} + \frac {\log \left (x - \sqrt {\frac {c^{2} d^{2} - 2 \, \sqrt {-a c} d^{3}}{d^{4}}} + \frac {c}{d}\right )}{4 \, {\left (d^{2} {\left (\sqrt {\frac {c^{2} d^{2} - 2 \, \sqrt {-a c} d^{3}}{d^{4}}} - \frac {c}{d}\right )}^{3} + 3 \, c d {\left (\sqrt {\frac {c^{2} d^{2} - 2 \, \sqrt {-a c} d^{3}}{d^{4}}} - \frac {c}{d}\right )}^{2} + 2 \, c^{2} {\left (\sqrt {\frac {c^{2} d^{2} - 2 \, \sqrt {-a c} d^{3}}{d^{4}}} - \frac {c}{d}\right )}\right )}} \]

[In]

integrate(1/(d^2*x^4+4*c*d*x^3+4*c^2*x^2+4*a*c),x, algorithm="giac")

[Out]

-1/4*log(x + sqrt((c^2*d^2 + 2*sqrt(-a*c)*d^3)/d^4) + c/d)/(d^2*(sqrt((c^2*d^2 + 2*sqrt(-a*c)*d^3)/d^4) + c/d)
^3 - 3*c*d*(sqrt((c^2*d^2 + 2*sqrt(-a*c)*d^3)/d^4) + c/d)^2 + 2*c^2*(sqrt((c^2*d^2 + 2*sqrt(-a*c)*d^3)/d^4) +
c/d)) + 1/4*log(x - sqrt((c^2*d^2 + 2*sqrt(-a*c)*d^3)/d^4) + c/d)/(d^2*(sqrt((c^2*d^2 + 2*sqrt(-a*c)*d^3)/d^4)
- c/d)^3 + 3*c*d*(sqrt((c^2*d^2 + 2*sqrt(-a*c)*d^3)/d^4) - c/d)^2 + 2*c^2*(sqrt((c^2*d^2 + 2*sqrt(-a*c)*d^3)/
d^4) - c/d)) - 1/4*log(x + sqrt((c^2*d^2 - 2*sqrt(-a*c)*d^3)/d^4) + c/d)/(d^2*(sqrt((c^2*d^2 - 2*sqrt(-a*c)*d^
3)/d^4) + c/d)^3 - 3*c*d*(sqrt((c^2*d^2 - 2*sqrt(-a*c)*d^3)/d^4) + c/d)^2 + 2*c^2*(sqrt((c^2*d^2 - 2*sqrt(-a*c
)*d^3)/d^4) + c/d)) + 1/4*log(x - sqrt((c^2*d^2 - 2*sqrt(-a*c)*d^3)/d^4) + c/d)/(d^2*(sqrt((c^2*d^2 - 2*sqrt(-
a*c)*d^3)/d^4) - c/d)^3 + 3*c*d*(sqrt((c^2*d^2 - 2*sqrt(-a*c)*d^3)/d^4) - c/d)^2 + 2*c^2*(sqrt((c^2*d^2 - 2*sq
rt(-a*c)*d^3)/d^4) - c/d))

Mupad [B] (verification not implemented)

Time = 12.19 (sec) , antiderivative size = 1551, normalized size of antiderivative = 2.93 \[ \int \frac {1}{4 a c+4 c^2 x^2+4 c d x^3+d^2 x^4} \, dx=\text {Too large to display} \]

[In]

int(1/(4*a*c + 4*c^2*x^2 + d^2*x^4 + 4*c*d*x^3),x)

[Out]

atan(((-(2*d*(-a^3*c^3)^(1/2) + a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2)*(((256*a*c^4*d^5 + 256*a*c^3*d^6*
x)*(-(2*d*(-a^3*c^3)^(1/2) + a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) - 64*a*c*d^6)*(-(2*d*(-a^3*c^3)^(1/2
) + a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) + 4*c*d^5 + 4*d^6*x)*1i + (-(2*d*(-a^3*c^3)^(1/2) + a*c^3)/(6
4*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2)*(((256*a*c^4*d^5 + 256*a*c^3*d^6*x)*(-(2*d*(-a^3*c^3)^(1/2) + a*c^3)/(64*(
a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) + 64*a*c*d^6)*(-(2*d*(-a^3*c^3)^(1/2) + a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2))
)^(1/2) + 4*c*d^5 + 4*d^6*x)*1i)/((-(2*d*(-a^3*c^3)^(1/2) + a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2)*(((25
6*a*c^4*d^5 + 256*a*c^3*d^6*x)*(-(2*d*(-a^3*c^3)^(1/2) + a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) - 64*a*c
*d^6)*(-(2*d*(-a^3*c^3)^(1/2) + a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) + 4*c*d^5 + 4*d^6*x) - (-(2*d*(-a
^3*c^3)^(1/2) + a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2)*(((256*a*c^4*d^5 + 256*a*c^3*d^6*x)*(-(2*d*(-a^3*
c^3)^(1/2) + a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) + 64*a*c*d^6)*(-(2*d*(-a^3*c^3)^(1/2) + a*c^3)/(64*(
a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) + 4*c*d^5 + 4*d^6*x)))*(-(2*d*(-a^3*c^3)^(1/2) + a*c^3)/(64*(a^2*c^6 + 4*a^3*
c^3*d^2)))^(1/2)*2i + atan((((2*d*(-a^3*c^3)^(1/2) - a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2)*(((256*a*c^4
*d^5 + 256*a*c^3*d^6*x)*((2*d*(-a^3*c^3)^(1/2) - a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) - 64*a*c*d^6)*((
2*d*(-a^3*c^3)^(1/2) - a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) + 4*c*d^5 + 4*d^6*x)*1i + ((2*d*(-a^3*c^3)
^(1/2) - a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2)*(((256*a*c^4*d^5 + 256*a*c^3*d^6*x)*((2*d*(-a^3*c^3)^(1/
2) - a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) + 64*a*c*d^6)*((2*d*(-a^3*c^3)^(1/2) - a*c^3)/(64*(a^2*c^6 +
4*a^3*c^3*d^2)))^(1/2) + 4*c*d^5 + 4*d^6*x)*1i)/(((2*d*(-a^3*c^3)^(1/2) - a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2
)))^(1/2)*(((256*a*c^4*d^5 + 256*a*c^3*d^6*x)*((2*d*(-a^3*c^3)^(1/2) - a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^
(1/2) - 64*a*c*d^6)*((2*d*(-a^3*c^3)^(1/2) - a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) + 4*c*d^5 + 4*d^6*x)
- ((2*d*(-a^3*c^3)^(1/2) - a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2)*(((256*a*c^4*d^5 + 256*a*c^3*d^6*x)*(
(2*d*(-a^3*c^3)^(1/2) - a*c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) + 64*a*c*d^6)*((2*d*(-a^3*c^3)^(1/2) - a*
c^3)/(64*(a^2*c^6 + 4*a^3*c^3*d^2)))^(1/2) + 4*c*d^5 + 4*d^6*x)))*((2*d*(-a^3*c^3)^(1/2) - a*c^3)/(64*(a^2*c^6
+ 4*a^3*c^3*d^2)))^(1/2)*2i